A secant line intersects the graph of $g(x)=4x^2-5$ at two points with $x$ -coordinates $-4$ and $-4+h$, where $h\neq0$. What is the slope of the secant line in terms of $h$ ? Your answer must be fully expanded and simplified.
Answer: We are given that the secant line intersects the graph at $x=-4$ and $x=-4+h$. Since these points are on the the graph of $g(x)=4x^2-5$, we know that they must be $(-4,59)$ and $(-4+h,\,4(-4+h)^2-5)$, respectively. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{4(-4+h)^2-5-59}{-4+h-(-4)} \\\\ &=\dfrac{4(-4+h)^2-64}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{4(-4+h)^2-64}{h} \\\\ &=\dfrac{4(16-8h+h^2)-64}{h} \\\\ &=\dfrac{64-32h+4h^2-64}{h} \\\\ &=\dfrac{4h^2-32h}{h} \\\\ &=\dfrac{h(4h-32)}{h} \\\\ &=4h-32\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, we can conclude that the slope of the secant line is $4h-32$.